JEE Main & Advanced
JEE Main Paper (Held on 09-4-2019 Afternoon)
question_answer
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :- [JEE Main 9-4-2019 Afternoon]
A)\[\frac{V}{K+n}\]
B)(v)\[V\]
C)\[\frac{(n+1)V}{(K+n)}\]
D)\[\frac{nV}{K+n}\]
Correct Answer:
C
Solution :
After fully charging, battery is disconnected Total charge of the system \[=CV+nCV\] \[=(n+1)CV\] After the insertion of dielectric of constant K New potential (common) \[{{V}_{C}}=\frac{total\,ch\arg e}{total\,capaci\,\tan ce}\] \[=\frac{\left( n+1 \right)CV}{KC+nC}=\frac{(n+1)V}{K+n}.\]